AERSP 301
Shear of beams
(Open Cross

section)
Jose Palacios
Shear of Open and Closed Section Beams
•
Megson
–
Ch. 17
•
Open Section Beams
–
Consider only shear loads applied through shear center (no twisting)
–
Torsion loads must be considered separately
•
Assumptions
–
Axial constraints are negligible
–
Shear stresses normal to beam surface are negligible
•
Near surface shear stress = 0
•
Walls are thin
–
Direct and shear stresses on planes normal to the beam surface are const through
the thickness
–
Beam is of uniform section
•
Thickness may vary around c/s but not along the beam
–
Thin

Walled
•
Neglect higher order terms of t (t
2
, t
3
, …)
•
Closed Section Beams
–
Consider both shear and torsion loading
Force equilibrium: General stress, Strain, and Displacement
Relationships
•
S
–
the distance measured
around the c/s from some
convenient origin
•
σ
z
–
Direct stress
(due to
bending moments or bending
action of shear loads)
•
–
Shear stresses due to
shear loads or torsion loads
(for closed section)
•
σ
s
–
Hoop stress, usually zero
(non

zero due to internal
pressure in closed section
beams)
zs
=
sz
=
shear flow; shear
force per unit length
q =
t
(positive in the direction of s)
Force equilibrium (cont’d)
•
From force equilibrium considerations in z

direction:
•
Force equilibrium in s

direction gives
Stress Strain Relationships
•
Direct stress:
z
and
s
strains
z
and
s
•
Shear stress:
strains
(=
zs
=
zs
)
•
Express strains in terms of displacements of a
point on the c/s wall
•
v
t
and v
n
: tangential and normal
displacements in xy plane
Not used
(1/r: curvature of wall in x

y plane)
Stress Strain Relationships
•
To obtain the shear strain, consider the element below:
•
Shear strain:
Center of Twist
•
Equivalent to pure rotation about some pt. R
(center of twist [for loading such as pure torsion])
•
For the point N
•
Origin O of axes
chosen arbitrarily,
and axes undergo
disp.
u, v,
Center of Twist (cont’d)
•
But
Equivalent to pure rotation about some pt. R
(center of twist [for loading such as pure torsion])
Center of twist cont…
•
Also from
•
Comparing Coefficients with:
Position of Center of Twist
Shear of Open Section Beams
•
The open section beam supports
shear loads S
x
and S
y
such that there
is no twisting of the c/s (i.e. no torsion
loads)
•
For this, shear loads must pass
through a point in the c/s called the
SHEAR CENTER
–
Not necessarily on a c/s member
•
Use the equilibrium eqn.
•
And obtaining
z
from basic bending theory
(no hoop stresses,
s
= 0)
Shear of Open Section Beams cont…
From:
Shear of Open Section Beams cont…
•
Integrating with respect to s starting from an origin at an
open edge (q = 0 at s = 0) gives:
•
For a c/s having an axis of symmetry, I
xy
= 0. Then eq.
for q
s
simplifies to:
Shear sample problem
•
Determine the shear flow distribution
in the thin

walled z

section shown
due to shear load
S
y
applied through
its shear center (no torsion).
•
Where is the shear center?
•
And the centroid?
t
y
x
h
h/2
1
2
4
3
Shear Flow Distribution (S
x
= 0):
s
xy
yy
xx
yy
y
s
xy
yy
xx
xy
y
s
yds
t
I
I
I
I
S
xds
t
I
I
I
I
S
q
0
2
0
2
Shear sample problem
8
12
3
3
3
3
th
I
th
I
th
I
xy
yy
xx
ds
y
x
h
S
yds
t
I
xds
t
I
I
I
I
S
q
s
y
s
yy
s
xy
xy
yy
xx
y
s
)
48
.
6
32
.
10
(
0
3
0
0
2
Bottom Flange: 1

2,
y =

h/2, x =

h/2 + S
1
0
≤ S
1
≤ h/2
ds
h
S
h
h
S
q
ds
y
x
h
S
q
s
y
s
y
s
))
2
/
(
48
.
6
)
2
/
(
32
.
10
(
)
48
.
6
32
.
10
(
1
0
1
3
12
0
3
Show this: EXAM TYPE PROBLEM
Shear sample problem
h
S
q
)
h
(S
q
S
hS
S
h
S
ds
h
S
h
h
S
q
y
y
s
y
42
.
0
2
/
2
@
0
)
0
(
1
@
)
74
.
1
16
.
5
(
))
2
/
(
48
.
6
)
2
/
(
32
.
10
(
2
1
1
1
1
2
1
3
0
1
3
12
1
t
y
x
h
h/2
1
2
4
3
Shear sample problem
2
2
2
2
3
2
2
2
0
3
23
42
.
3
42
.
3
42
.
0
)
(
48
.
6
)
(
42
.
3
(
2
s
hs
h
h
S
q
ds
s
h
h
S
q
y
s
y
In web 2

3:
y =

h/2 + S
2
x = 0 for 0
≤
S
2
≤
h
Symmetric distribution about
C
x
with max value at
S
2
= h/2 (y = 0)
and positive shear flow along the web
Shear Flow
S
2
= 0
h
S
q
y
42
.
0
2
Shear sample problem
In web 3

4:
y =h/2
x = S
3
for 0
≤
S
3
≤
h
3
3
0
3
3
34
)
(
24
.
3
)
(
32
.
10
(
3
q
ds
h
s
h
S
q
s
y
Shear Flow Distribution in z

section
Calculation of Shear Center
•
If a shear load passes
through the shear center,
it will produce NO TWIST
–
M = 0
•
If c/s has an axis of
symmetry, the shear
center lies on this axis
•
For cruciform or angle
sections, the shear center
is located at the
intersection of the sides
Sample Problem
•
Calculate the shear
center of the thin

walled
channel shown here:
Sample problem shear center
The shear center (point S) lies on the horizontal
(C
x
)
axis
at some distance
ξ
s
from the web
. If a shear load
S
y
passes through
the shear center it will produce
no twist.
Let’s look at the shear flow distribution due to S
y
:
Since
I
xy
= 0
and
S
x
= 0
S
xx
y
s
ds
ty
I
S
q
0
)
6
1
(
12
3
h
b
th
I
xx
Further:
S
y
s
ds
y
h
b
h
S
q
0
3
6
1
12
Then:
Sample problem shear center
Along the bottom flange 1

2,
y =

h/2
1
2
0
1
3
12
6
1
6
2
/
6
1
12
1
s
h
b
h
S
ds
h
h
b
h
S
q
y
S
y
At point 2:
S
1
= b
b
h
b
h
S
q
y
6
1
6
2
2
Along the web 2

3 ,
y =

h/2 + S
2
Sample problem shear center
b
h
b
h
S
s
s
h
h
b
h
S
q
ds
s
h
h
b
h
S
q
y
y
S
y
6
1
6
2
2
6
1
12
2
/
6
1
12
2
2
2
2
3
2
0
2
2
3
23
2
At point 3:
S
1
= h
2
2
3
6
1
6
q
b
h
b
h
S
q
y
Along the top flange 3

4 ,
y = h/2
Sample problem shear center
b
h
b
h
S
s
h
h
b
h
S
q
ds
h
h
b
h
S
q
y
y
S
y
6
1
6
2
6
1
12
2
/
6
1
12
2
3
3
3
0
3
3
34
3
At point 4:
S
3
= b
0
4
q
Good Check!
Sample problem shear center
Shear Flow Distribution due to
S
y
The moments due to this shear flow
distribution should be equal to zero about the shear center
2
0
0
23
1
12
2
2
ds
q
ds
h
q
s
b
h
S
ξ
s
S
y
Solve for
ξ
s
to find the shear center location:
s
h
y
b
h
s
y
y
ds
b
h
b
h
S
ds
s
s
h
h
b
h
S
ds
s
h
h
b
h
S
0
2
2
0
0
2
2
2
2
3
1
1
2
6
1
6
2
2
6
1
12
2
6
1
6
2
h
b
h
b
s
1
3
2
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